Prove $\forall a,b \in \mathbb{Z}, 18a + 6b \ne 1$

Question

Prove $\forall a,b \in \mathbb{Z}, 18a + 6b \ne 1$

Is there a way to do this using proof by contradiction without using mod?

Answer 1

It is quite easy; suppose that there exist integers $a$ and $b$ such that $18a+6b=1$, then $6(3a+b)=1$, and that implies that $3a+b$ (which is an integer for any $a$ and $b$) is an multiplicative inverse of $6$. This is a contradiction, because the only integers with inverses are $1$ and $-1$.

Answer 2

Hint: $$18a+6b=6\cdot(3a+b)$$